A boy can throw a stone upto a maximum height of 10 m. The maximum horizontal distance that the boy

Q: Why is the maximum height achieved when the stone is thrown vertically at 90 degrees?

The vertical component of velocity is $u \sin\theta$. To maximize the height, you need to maximize this vertical component. Since the maximum value of $\sin\theta$ is 1 (at $\theta = 90^\circ$), throwing it straight up gives the maximum possible height.

Q: Why does an angle of 45 degrees give the maximum horizontal range?

The formula for range is $R = \frac{u^2 \sin(2\theta)}{g}$. To get the maximum range, the term $\sin(2\theta)$ must be at its maximum value, which is 1. This happens when $2\theta = 90^\circ$, meaning $\theta = 45^\circ$.

Q: Do I need to know the exact value of gravity (g) to solve this problem?

No, you don't! As you can see in the solution, the 'g' from the $u^2$ term perfectly cancels out with the 'g' in the denominator of the range formula. This makes the relationship $R_{\max} = 2H_{\max}$ independent of the planet you are on!

Q: Can I calculate the exact initial velocity of the boy's throw?

Yes, if you assume a value for gravity. If we take $g = 10\text{ m/s}^2$, then $u^2 = 20 \times 10 = 200$. So, the initial velocity $u$ would be $\sqrt{200} = 10\sqrt{2}\text{ m/s}$.

Q: Is the maximum horizontal range always exactly twice the maximum vertical height?

Yes, for a given initial speed $u$. The maximum height is $H_{\max} = \frac{u^2}{2g}$ (thrown at $90^\circ$), and the maximum range is $R_{\max} = \frac{u^2}{g}$ (thrown at $45^\circ$). By comparing the two, it is mathematically guaranteed that $R_{\max} = 2H_{\max}$.

Question

A boy can throw a stone upto a maximum height of $10	ext{ m}$. The maximum horizontal distance that the boy can throw the same stone upto will be

Options:

(A) $20\sqrt{2}	ext{ m}$

(B) $10	ext{ m}$

(C) $10\sqrt{2}	ext{ m}$

(D) $20	ext{ m}$

Sigma Percentile · Accepted Answer

### The Human Limit: Throwing a Stone

Imagine you are standing in an open field with a stone in your hand. You want to test your absolute physical limits. First, you decide to throw the stone straight up into the sky as hard as you can. This vertical throw represents your maximum power output, translating into a specific initial velocity, let's call it $u$.

The height the stone reaches is entirely dependent on this initial velocity. Once the stone leaves your hand, gravity takes over, decelerating it at a rate of $g$ until it momentarily stops at its peak. This is the classic scenario of one-dimensional kinematics.

### Analyzing the Vertical Setup

When you throw the stone vertically upwards, the angle of projection is $90^\circ$ with respect to the horizontal. Using the third equation of motion, $v^2 = u^2 - 2gh$, we can analyze the journey to the peak. At the maximum height, the final velocity $v$ becomes zero.

$$0 = u^2 - 2gH_{\max}$$

Rearranging this, we get the formula for the maximum vertical height:

$$H_{\max} = \frac{u^2}{2g}$$

The problem states that the boy can throw the stone up to a maximum height of $10	ext{ m}$. By substituting this value into our equation, we can extract the raw power of the boy's throw, represented by $u^2$:

$$10 = \frac{u^2}{2g} \implies u^2 = 20g$$

Notice how we didn't bother calculating the exact value of $u$. In physics, it is often much more elegant to carry forward intermediate expressions like $u^2$ rather than evaluating them prematurely. This prevents rounding errors and often leads to beautiful cancellations later on.

### The Quest for Maximum Range

Now, you change your goal. Instead of throwing the stone high, you want to throw it as far away as possible across the field. You are now dealing with two-dimensional projectile motion. The horizontal distance covered by the stone is called its Range ($R$).

The formula for the horizontal range of a projectile is given by:

$$R = \frac{u^2 \sin(2	heta)}{g}$$

To maximize this range for a given initial velocity $u$, we need to maximize the trigonometric term $\sin(2	heta)$. The maximum value of the sine function is $1$, which occurs when the angle is $90^\circ$. Therefore, $2	heta = 90^\circ$, which means $	heta = 45^\circ$.

When you throw the stone at exactly $45^\circ$, the range formula simplifies beautifully to:

$$R_{\max} = \frac{u^2}{g}$$

### The Final Calculation

We have arrived at the final stage of our problem. We know the formula for the maximum horizontal range, and we already found the value of $u^2$ from the vertical throw. It's time to bring them together.

Substitute $u^2 = 20g$ into the maximum range equation:

$$R_{\max} = \frac{20g}{g}$$

The acceleration due to gravity, $g$, cancels out perfectly from the numerator and the denominator. This is the elegance of physics at play!

**$$R_{\max} = 20	ext{ m}$$**

This reveals a fascinating and highly useful universal relationship: For any given initial speed, the maximum horizontal range is exactly twice the maximum vertical height ($R_{\max} = 2H_{\max}$). If you can throw a ball $10	ext{ m}$ high, you can throw it $20	ext{ m}$ far. Keep this golden rule in your mental toolkit for quick problem-solving!

Animated Solution for Physics - Kinematics: A boy can throw a stone upto a maximum height of $10 m$ . The maximum horizontal distance that the boy can throw the same stone upto will be

Select Answer:

Visualized Solution

\text{Visualizing the Maximum Height}

\text{Extracting Initial Velocity}

\text{Condition for Maximum Range}

\text{Formula for Maximum Range}

\text{Calculating Maximum Range}

\text{Conclusion}

The Sigma Insight: Projectile Motion

The Human Limit

Analyzing the Vertical Setup

The Quest for Maximum Range

The Final Calculation

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